The long-term average number of support staff (N) in a stable system is equal to the total number of customers in the queue (Q) multiplied by the average time support staff spend resolving a customers’ technical problems (T), divided by the total time waited in the queue (W); or expressed algebraically: N=QT/W.

Thinking things through a bit more, the total number of customers in the queue (Q) at a point in time in a stable system should be equal to the rate at which people joined the queue (λ), minus the rate at which the support desk dealt with technical problems (i.e. N/T) over the period of observation. Obviously Q>=0.

So N=QT/W and Q=λ-N/T which all comes out in the wash as:

N=λT/(W+1)

I thought might be a bit of fun to see if this could be applied to the customer support call centre waiting statistics published by one of Australia’s largest ISPs, iiNet.

iiNet make some support centre data available via their customer toolbox page. Below is a screenshot of call activity and wait times graphed each hour by iiNet on 10 January 2012. The green line (in conjunction with the scale on the left hand side of the graph) represents the average time (in minutes) it took to speak to a customer service representative (CSR), including call-backs. The grey bars (in conjunction with the right hand scale) represents the total number of incoming phone calls to iiNet’s support desk.

It may be possible to use the formula derived above to estimate how many CSRs iiNet had on the support desk handling calls that day. For example, during the observed peak period of 8am to 1pm on Tuesday, 10 January 2012, the iiNet support desk was getting around 732 calls per hour on average. The expected wait time in the queue over the same period was around 11 minutes.

If we assume that the average time taken for a CSR to resolve a technical problem is, let’s say, 12.5 minutes, then we can estimate that the number of CSRs answering calls in a typical peak-hour between 8am to 1pm on 10 January 2012 as:

732*(12.5/60) / (11/60 + 1)

= 129 CSRs actively handling calls.

Sounds sort of a reasonable for a customer service-focussed ISP the size of iiNet. But if iiNet wanted to bring the average time in the queue down even more – to a more reasonable 3 minutes, for example – they’d need 145 CSRs (all else remaining equal) during a typical peak-hour answering calls.

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I updated my “Stock Market Seismometer” (click on the separate tab above for details) for the first time in many months. I have to say, the results shocked me. Over the course of 2011 the Australian stock market slid down even further into “oversold” territory. As we head into 2012 things have never looked so bleak. Or maybe 2012 will be the year of the rebound? I expect at some point growth will move back to its long term trend, but when that will start to happen is anyone’s guess.

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Poker machines are unique in the gambling world. They are the only form of gambling that has been designed and crafted for the purpose of making money, and where there is absolutely no chance of influencing the outcome.

- Tom Cummings, Poker machine maths, 27 May 2011

The article, linked above, on poker machines is an excellent insight into the mathematics of poker machine gaming and how, despite a guaranteed return-to-player percentage which is high, a punter that plays long enough ends up with nothing. The vital point that Tom Cummings makes is this:

It’s common knowledge that poker machines have a return-to-player percentage of anywhere between 85 per cent and 90 per cent, depending on where you live and what kind of establishment you’re playing in. For the sake of the story, let’s assume that Gladys’ poker machine was set to 90 per cent. That means that over a long period of time, the machine will return 90 per cent of money gambled to players, and keep 10 per cent as profit.

But wait, I hear you cry. Gladys didn’t lose 10 per cent; she lost it all! Well, no… not according to poker machine mathematics. The rule is that the poker machine has to return 90 per cent of money gambled… not money inserted. And there’s a huge difference.

The way it works, mathematically, is, I think, very interesting. Take the example used in the article of a 2-cent gaming machine, at $1 per play, with guaranteed return of 90%. When you load that dollar in the slot, sometimes you win, and sometimes you lose, but the long-term average dictates that at each “turn” the dollar is losing 10% of its value. $1 turns into 90c, turns into 81c, turns into 72.9c, … , turns into 2c at which point the game ends.

More generally, at “turn” n, an amount A₁ initially fed into a slot machine with return R is worth:

or solving for n:

What does this mean? It means that it takes, on average

n = [ ln(0.02) - ln(1.00) / ln(0.90) ] + 1 = 38

iterations to turn $1.00 into 2 cents on a 90% return poker machine. More importantly, the total amount gambled isn’t $1.00, it’s actually (because you’re re-investing your winnings and following your losses until the game ends): $1.00 + 90c + 81c + 72.9c + … + 2c. Or more generally,

which you’ll remember converges to:

So a single dollar coin will generate, on average (on a 2-cent, 90% return machine):

[ 1-0.9^(38-1)] / [ 1-0.90 ] = $9.80 in total bets.

At $1.00 per game, at 10 games per minute, that’s about 1 minute of play for every dollar put in, or 5 hours to burn through Gladys’ $300.

Those who defend poker machines often point to the high rate of return as one of the reasons that pokies are just “good, clean fun” for most people. The reality is that every poker machine can meet this “rate of return” requirement while still leaving the gambler broke. That’s poker machine mathematics.

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Further reading:

ABC Hungry Beast, The Beast File: Pokies

Australia’s 43rd parliament is a precarious one. Last year’s federal election resulted in a hung parliament in the House of Representatives, with a minority Labor government clinging to power with the help of the Greens and a few Independents. Now I don’t want to sound morbid but, Heaven forbid, should any one of the Honorable Members die, the resulting by-election could conceivably bring down the government. So finely balanced are the numbers.

So, I wonder, what IS the probability that this will happen before the next election, due in 2013?

Statisticians that devote themselves to thinking about exactly this sort of question are called Actuaries, and when calculating risk of mortality they reference a thing called a Life Table. Life tables list the probabilities that a person aged x will survive to x+1 years. For example, looking at the most recent Life Tables for Australia, the probability that a female aged 40 will die before turning 41 is 0.00078 (i.e. 0.078%). Or looking at it in a more positive light, the chance a 40 year old female will live to see her 41st birthday is 1-0.00078, or a healthy 99.922%.

So turning our attention back to the 150-member House of Representatives, what is the probability that one or more of them might shuffle off this mortal coil in the next two years, resulting in a by-election and/or a change of government?

To be rigorous I should aggregate individual probabilities of survival based on each member’s age and gender. But I can’t be arsed, so I’ll talk about it in terms of generalities.

If I recall correctly, the average age of a politician is 51 years. Despite this being a modern society in the year 2011, our parliament is still a massive sausage-fest. Men in government significantly outnumber women. So I’ll just be lazy and use the qx column in the Australian Life Tables linked above for males aged 51 and 52.

From the Life Table, the probability that a male aged 51 years will survive to his 53rd birthday is (1-0.00332)*(1-0.00359)=0.993. Therefore the chance that every member of the 150 seat parliament survives for the next two years can be approximated at 0.993^150=0.349 (i.e. 34.9%). The likelihood that one or more parliamentarian dies within the next two years is the complement all surviving, calculated as 1-0.349=0.651 (65.1%).

65%. That’s a pretty high risk that the parliament won’t see out its full term.

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The above headline and corresponding story appeared in ARN on 4 May 2011. The article was reporting on the Telecommunications Industry Ombudsman’s (TIO) January to March 2011 statistics published here.

Now before I go any further, let me say I have grave concerns with the source data itself, let alone the way it’s being reported in the press. When talking about TIO complaint statistics, it is worthwhile to familiarise yourself with this recent research report from the University of Technology, Sydney (UTS).

Findings of the UTS research include (from the executive summary):

• Over 80% (of carriage service providers surveyed) assert that the TIO accepts complaints which are out of jurisdiction, frivolous or vexatious.

Recommendations include:

• Amend TIO policy and procedure to cease the multiple counting of complaints in statistics and recommence reporting disposition of complaints.
• Amend TIO policy and procedure to refer to Level 1 Complaints as Contacts rather than Complaints.

So fair to say the TIO isn’t exactly highly regarded by the ISP industry. Indeed, Exetel have commenced legal action against the TIO for acting outside its charter. Or, in Exetel CEO John Linton’s own words, “Exetel is now going to take a court action to have the TIO closed as being a ‘criminal’ organisation based on a level of incompetence, lack of knowledge and unconstitutional actions that even Australians, who are mostly as apathetic as the governments they tolerate, shouldn’t have imposed on them.”

Well OK then…

Now if you still want to accept that all “complaints”, as reported by the TIO, are real, legitimate complaints then the fairest way to present the numbers is as a proportion of overall customer base. That’s going to be a bit tricky, because not all ISPs publish accurate customer numbers in the public domain. So I’ve had to estimate relative market share using Roy Morgan survey data and a methodology used in a previous blog post. When I refer to market share I mean ALL business, government AND home subscribers with ANY kind of internet access including dialup, DSL, cable, fibre, satellite and wireless (fixed & mobile). The Roy Morgan surveys were conducted between January 2010 and December 2010

Then, to be consistent, I only looked at TIO complaints made against a provider’s internet services . That is, I excluded complaints made against landline and mobile phone services. I considered all internet services TIO complaints made between October 2010 and March 2011

Then I derived a normalised score by dividing the relative market share by the proportion of complaints. The benchmark is a score of 1000 and a higher score is better. A score over 1000 means that an ISP recorded relatively fewer complaints as a proportion of its customer base than its peers. Less than 1000 means that an ISP was over-represented in TIO complaints relative to its size.

The TIO complaints leaderboard comes out as follows (click to up-size):

Table 1: TIO internet services complaints leaderboard, 2010/2011

iiNet did indeed do very well. With roughly an 11% market share, but only 6% of complaints, it gets a score of 1843. That’s well above the expected benchmark of 1000. But it didn’t fare best. The ISP that recorded the lowest number of TIO complaints as a proportion of number of customers was Internode, with a massive normalised score of 3448.

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CVC is bandwidth that ISPs will need to purchase from NBN Co, charged at the rate of $20 (ex-GST) per Mbps per month. Note that this CVC is on top of the backhaul and international transit required to pipe all those interwebs into your home. And just like backhaul and international transit, if an ISP doesn’t buy enough CVC from NBN Co to cover peak utilisation, its customers will experience a congested service.

The problem with the CVC tax, priced as it is by NBN Co, is that it punishes small players. By my calculations, an ISP of (say) 1000 subscribers will need to spend proportionally a lot more on CVC than an ISP of 1,000,000 subscribers if they want to provide a service that delivers the speeds it promises.

Here comes the statistics.

Consider NBN Co’s theoretical 12 megabit service with 10GB of monthly quota example that they use in the document I linked to above. 10GB per month, at 12Mbps gives you 6,827 seconds (a bit under 2 hours) at full speed before you’re throttled off. There’s 2,592,000 seconds in a 30-day month, so if I choose a random moment in time there is a 6827/2592000 = 0.263% chance that I’ll find you downloading at full speed.

That’s on average. The probability would naturally be higher during peak times. But let’s assume in this example that our theoretical ISP has a perfectly balanced network profile (no peak or off-peak periods). It doesn’t affect the point I’ll labour to make.

A mid-sized ISP with (let’s say) 100,000 subscribers can expect, on average, to have 100,000*0.263% = only 263 of those customers downloading at full speed simultaneously at any particular second. However, the Binomial distribution tells us that there’s a relatively small, but still statistically significant (at the alpha=0.05 level) probability that there could be 290 or more customers downloading at the same time.

So a quality ISP of 100,000 subscribers will plan to buy enough CVC bandwidth to service 263 customers at any one time. But a statistician would advise the ISP to buy enough CVC bandwidth to service 290 subscribers, an additional (290-263)/263 = 10% , or find itself with a congested service about one day in every 20.

This additional “burst headroom”, as a percentage, increases as the size of the ISP decreases. From above, an ISP of 10,000 can expect to have 26 customers downloading simultaneously at any random moment in time. But there’s a statistically significant chance this could be 35+. This requires them to buy an additional (35-26)/26=33% in CVC over and above what was expected to cover peak bursts.

The table below summarises, for ISPs of various sizes, how much additional CVC would need to be purchased over and above the expected amount, to provide an uncontended service 95%+ of the time.

Graphically it looks a bit like this…

As you can see, things only really start to settle down for ISPs larger than 100,000 subscribers. Any smaller than that and your relative cost of CVC per subscriber per month is disproportionally large.

….

Further reading:

Rebalancing NBNCo Pricing Model

NBN Pricing – Background & Examples, Part 1

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According to the latest Roy Morgan Internet Satisfaction data, Internode (93.4%) is still the top performer for customer satisfaction while iiNet (89.9%) appears to be closing the gap from 5.6% points in the 6 months to April 2010 to 3.5% points in the 6 months to May 2010.

Scrolling further down the Roy Morgan press release page, you’ll find individual ISP customer profiles available for purchase.  At the bottom of each report’s synopsis, you’ll see that a sample size has been included [for example, the Internode customer profile is based on a sample of 305 customers].  Now, combining these sample sizes from the ISPs’ profiles, along with the total number of internet subscribers figure from the Australian Bureau of Statistics, I think that could potentially provide a good basis for estimating market share.

My results/estimates are presented below, along with margins of error [95% confidence levels].  The estimate of total customers is ALL business, government and home subscribers with ANY kind of internet access including dialup, DSL, cable, fibre, satellite and wireless [fixed & mobile].  The Roy Morgan samples were taken between April 2009 and May 2010.  The ABS total internet subscribers figure is as at December 2009.

Table 1: Estimated Australian ISP market share, 2009-2010

• * Roy Morgan, April 2009 to May 2010
• ** Australian Bureau of Statistics, December 2009
• 95% CIs using method of Agresti and Coull [c.f. Engineering Statistics Handbook]
• AAPT, Netspace and Westnet are owned by iiNet
• Chariot is owned by TPG
• Adam offers residental internet access in South Australia & Northern Territory only

A [loud] word of warning: small sample sizes result in very wide, almost unusable margins of error.  For example, the 95% confidence interval around the Vodafone estimate is 0.1% [12,370 customers] to 5.1% [464,027 customers].  That’s a massive range for the true value to fall within, so be aware of these margins of error when perusing the table above.

Still, some interesting results, I think.

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So thank you to whoever nominated me.

I’ve updated my Stock Market Seismometer (see separate page tab above) to bring the data up to the end of the Australian 2009-10 financial year.  Not good news for share market punters.  Half of the recovery recorded throughout most of a hopeful 2009 calendar year has been wiped out.  How much further it has to fall, or whether it can turn things around, I just don’t know.  I don’t think anyone does.

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My objective, as it was last year, was to take results from these three survey questions related to customer service

1. When calling customer support, how long did you have to wait on the phone (or talk to an operator) before you spoke to the right person?
2. How quickly have technical support issues typically taken to resolve?
   
3. How would you rate their customer service?

and distil them down to a single score that can be used to rank providers.  I arbitrarily set the benchmark score across the whole industry to be 1000, with each individual ISP’s customer service ranked relative to that benchmark.  So an ISP score higher than 1000 is above the industry average.  Lower than 1000 is below average.

The methodology employed was exactly the same as last year, so no need to go into the details.  Without further ado, here are the updated results:

Stan’s Top Five ISPs for Customer Service in 2009 [2008 rank in brackets]

1. Adam Internet [3]
2. Westnet [1]
3. Amnet [2]
4. Internode [4]
5. iiNet [6]

Congratulations go to local Adelaide-based outfit, Adam Internet.  Number 1 with a bullet in 2009.  Westnet (purchased by iiNet in 2008) has always prided itself on providing subscribers with a premium customer service experience, so it was very surprising to see them knocked off their coveted number 1 spot.  Also surprising to see aaNet slip out of the Top Five altogether, replaced by iiNet.

The overall results from the three customer service questions (equally weighted) are as follows:

Table 1: Australian ISP customer service scores
<1000:below average … 1000:average … >1000:above average

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It’s important to keep in mind that Whirlpool’s Australian Broadband Survey isn’t scientific.  Although it gets tens of thousands of responses, it only reflects the opinions of those who are aware of the Whirlpool site and motivated to express an opinion.  It is a self-select survey and, as such, the respondents’ attitudes may not be statistically representative of the ISP’s customer base.   In other words, take with a grain of salt.

Ranked from highest to lowest the results are as follows:
ISP (2009 score) (2008 score):

• Adam Internet (1842) (1727)
• Westnet (1820) (2132)
• Amnet (1774) (1735)
• Internode (1488) (1348)
• iiNet (1284) (1081)
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• iPrimus (1066) (903)
• Exetel (991) (992)
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• aaNet (889) (1204)
• Netspace (777) (912)
• AAPT (755) (808)
• Optus Cable (748) (736)
• TPG (739) (963)
• Optus DSL (710) (672)
• Telstra Cable (586) (711)
• Telstra DSL (562) (676)
• Telstra NextG (437) (562)
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• Other (826) (959)
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• AVERAGE (1000)

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What I found intrigued me.  The Chinese population data did not conform to Benford’s Law.  I stress that this departure from what was expected should not be taken as any kind of proof of fraudulent data entry (Benford’s Law often applies, but not always).  However, it was a surprising result.

1. BENFORD’S LAW
Tally the number of times the first digit 1 through to 9 occurs in any “real world” data source such as lengths of rivers or heights of buildings (unit of measurement doesn’t matter).  Intuitively you would expect the digits should be uniformly distributed with each one being observed 1/9th (about 11%) of the time.  Under Benford’s Law, however, a first digit of 1 tends to appear with a frequency of about 30%, a leading digit of 2 tends to occur about 18% of the time, a 3 about 13%, and so on in a logarithmically decreasing pattern, with a leading digit of 9 often being observed less than 5% of the time.

2. DATA
Dan provided land area (in square kilometres), along with two series of population data: “actual” (常住人口); and official/registered (戶籍人口) for each of the more than 350 administrative areas in China.  These data can be viewed here.  Dan notes that at the bottom of the spreadsheet are four cities which the Chinese government counts as provinces.  These are included to bring the population to the official total for China.  If you scan through the columns of figures, you’ll see that some administrative areas don’t have a land area recorded next to them, or an “actual” population.  However, they all have a registered population.  These gaps mean that the totals presented in the tables below won’t match.  The data are as at the end of 2007.

3.1 BENFORD ANALYSIS – CHINESE LAND AREA
So on to the analysis.  Let’s start by tallying the occurrences of first digits in the column of figures holding the land area data.  Table 1 summarises these frequencies (number and percentage) that were observed and expected under Benford’s Law.

Table 1: Chinese administrative land areas (sq kms)

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I think the data are easier to digest when presented graphically.

Figure 1: Chinese administrative land areas (sq kms)

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The leading digits of Chinese land area data by administrative area do decrease roughly logarithmically, although don’t technically follow a Benford distribution when subject to a Chi-square goodness of fit test (X2=26.05 on 8 d.f.; p=0.00103).  Data are overweight in figures starting with a “1″ and underweight in areas leading with digits “3″ to “6″.  But as Benford Law predicts, “1″ is the most common leading digit, followed by “2″ with the remaining leading digits bringing up the rear.

3.2 BENFORD ANALYSIS – CHINESE ACTUAL POPULATION
Similarly, let’s look at the leading digits of “actual” populations by administrative areas.

Table 2: Chinese administrative area “actual” populations

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When plotted, the difference between frequencies observed and what was expected under Benford’s Law is, I think, striking.

Figure 2: Chinese administrative area “actual” populations

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Not even close to a Benford distribution.  The leading digit of “1″ actually occurs less often than a leading digit of “2″, although after that things do decrease logarithmically.

3.3 BENFORD ANALYSIS – CHINESE REGISTERED POPULATION
Dan asked that I focussed on the registered/official population figures (the highlighted column of raw data in the spreadsheet linked above).  Leading digits from the registered, official population figures by administrative areas are presented in Table 3 and Figure 3 below.

Table 3: Chinese administrative area “registered” populations

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Figure 3: Chinese administrative area “registered” populations

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Things look a bit better than the “actual” population data, but registered populations certainly don’t follow Benford’s Law either.  The leading digit of “1″ is severely under-weight compared to its predicted frequency and, in fact, digits “1″ through to “3″ from registered population data occur with almost equal frequency.

4. SUMMARY
Chinese land area data by Chinese government administrative area follow a logarithmically decreasing pattern, although technically not a Benford distribution.  Chinese population data (“actual” and registered) by the same areas don’t follow a logarithmically decreasing pattern at all, Benford or otherwise.

That these Chinese government data don’t conform to Benford’s Law should not be taken as any kind of proof or insinuation that something untoward is going on.  It’s something that I would say warrants further investigation, but I expect there’ll be a rational explanation for it.  It’s probably more indicative of my own gaps in understanding than anything else.  Having said that, it was a very interesting exercise and I’d like to thank Dan Silver for the opportunity to write about it.

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